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Compton Scattering Experiment

  1. Oct 17, 2004 #1
    Any ideas how to do this question ?

    In a Compton type experiment X-rays scattered through 90 degrees suffered one percent reduction in energy.What was the incident energy of the X-rays ? You can assume that the electron coiled non-relativistically.

    Any help would be really appreciated ! :smile:

    Bob
     
  2. jcsd
  3. Oct 17, 2004 #2
    Oh by the way,i used the equation :


    dL=h/mc (1-cos angle)

    and i got dL=2.43 E -12

    Since the energy reduction is 1 % i did

    hc/L-hc/L' =0.01

    and L'=L+2.43 E -12

    so we end up with a quadratic :

    L^2 - (2.43 E -12)L-(4.83 E -35)=0

    Solving I got, L= O or -2.43 E -12

    Hence incident energy is hc/0 or hc/-2.43 E -12

    => energy =infinity :surprised or hc/(-2.43 E -12). :confused:

    Now is this total rubbish,or am I getting somewhere ? :bugeye:

    Cheers for any help !
    Bob
     
  4. Oct 17, 2004 #3
    I believe you made an error when calculating the final energy (and hence wavelength) of the X-ray. Do it like this. The final energy of the X-ray [tex]E_f = 0.99E_i[/tex] where the i stands for "initial". Now use the Einstein relation for the energy of a photon: [tex]E = \frac{hc}{\lambda}[/tex] to figure out the relationship between the final and initial wavelengths of the X-rays. You shouldn't get any type of quadratic term in [tex]\lambda[/tex]
     
  5. Oct 17, 2004 #4
    Thanks mate ! that was exactly my mistake.I cant believe I made that error :cry: but oh well !. :rolleyes:

    Cheers again !

    This world needs more people like :approve: you
     
  6. Oct 31, 2004 #5

    Andrew Mason

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    I think the key is to get the recoil angle of the electron. It is almost 45 degrees ([itex]tan\theta = .99[/itex]). Plug that into the Compton formula and you will get the magnitude of the wavelength change. That change represents a loss of one percent of the original energy.

    AM
     
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